Question

. Find the work done by the vector field F~ (x, y, z) = 2y~i + e x~j − z 3~k moving a particle along the curve C given by ~r(t) = (t 2 , t2 + 2, t), 0 ≤ t ≤ 1.

Answer 1

$\tilde F(x,y,z)=2y\,\tilde\imath+e^x\,\tilde\jmath-z^3\,\tilde k$

$\implies\tilde F(\tilde r(t))=2(t^2+2)\,\tilde\imath+e^{t^2}\,\tilde\jmath-t^3\,\tilde k$

$\tilde r(t)=t^2\,\tilde\imath+(t^2+2)\,\tilde\jmath+t\,\tilde k$

$\implies\dfrac{\mathrm d\tilde r}{\mathrm dt}=2t\,\tilde\imath+2t\,\tilde\jmath+\tilde k$

The work done by $\tilde F$ along $C$ is

$\displaystyle\int_C\tilde F(x,y,z)\cdot\mathrm d\tilde r=\int_0^1\tilde F(\tilde r(t))\cdot\frac{\mathrm d\tilde r}{\mathrm dt}\,\mathrm dt=\int_0^1(3t^3+8t+2te^{t^2})\,\mathrm dt=\boxed{e+\frac{15}4}$

Answer 2

In this exercise we have to use the knowledge of vectors to calculate the field of the given vector, in this way we have that:

$\int\limits_C {F} \, dr = e+\frac{15}{4}$

From the given equation we find that:

$F(x, y, z) = 2y+e^x-z^3$

Substituting the value of r in the above equation, we have:

$F(r(t))=2(t^2+2)+e^{t^2}-t^3\\r(t)=t^2+(t^2+2)+t\\\frac{dr}{dt}= 2t+2t$

So the work of function F will be that:

$\int\limits_C {F(x, y, z)} \, dr= \int\limits_C {F(r(t))} \, dt = \int\limits^1_0 {(3t^3+8t+2te^{t^2}} \, dt= e+\frac{15}{4}$

See more about vectors at brainly.com/question/13188123